Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let X be an integral scheme that is separated (say over an affine scheme). Define a Weil divisor as a finite integral combination of height 1 points of X, where the height of a point of X is the dimension of its local ring. Let Z be a closed subscheme of X, Z not equal to X.

Is there an example of such an X and Z where Z contains infinitely many Weil divisors?

share|improve this question

3 Answers 3

up vote 12 down vote accepted

Yes, there are such examples (with X quasi-compact, otherwise it is trivial). There is a general result due to Hochster ("Prime Ideal structures in commutative rings", his thesis) which says that spectra of rings are exactly those topological spaces X such that:

1) X is Kolmogorov (T_0).

2) X is quasi-compact.

3) The quasi-compact open subsets form an open basis.

4) X is quasi-separated, i.e., quasi-compact open subsets are closed under finite intersections.

5) Every non-empty irreducible closed subset has a generic point.

We will construct an irreducible topological space X satisfying 1-5 with underlying set 2^N U {x} such that 2^N is closed in X and totally disconnected and X has generic point x. If X=Spec(A), then the closed subvariety 2^N is a union of infinitely many Weil divisors. (X has dimension 1)

The topology on 2^N will be the product topology, i.e., that of the 2-adic integers (or if you prefer, the Cantor set). An open basis for this topology are cylinders, i.e., sets where a finite number of components are fixed. The cylinders are also closed. The space 2^N is Hausdorff, compact and totally disconnected, hence satisfies 1-5.

An open basis for the topology of X = 2^N U {x} is given by sets of the form W U {x} where W is a cylinder or the empty set. The quasi-compact open subsets of X are the finite unions of such sets and X satisfies 1-4. To see 2-4 note that if V is an open subset of X then

V is quasi-compact <=> The intersection of V and 2^N is clopen

Finally, the non-empty irreducible closed subsets of X are the singleton sets of 2^N and X itself and these all admit generic points so X satisfies 5.

Remark: X seems to be closely related to the spectrum of the Tate-algebra Q_p<x>. The canonical topology on the closed points are exactly the p-adic integers. But the Zariski topology is completely different (Q_p<x> is noetherian and regular of dimension 1).

share|improve this answer

Let R be the integral closure of $\mathbb{Z}$ in the algebraic closure of $\mathbb{Q}$. Then $\dim (R)=1$ and there are infinitely many prime ideals lying above a prime ideal $p\mathbb{Z}$. Hence the closed set $V(pR)$ has the required property.

share|improve this answer

Take X to be A^\infty and Z to be the union of the coordinate hyperplanes.

Edit: Jonathan pointed out to me that the infinite union of hyperplanes in A^\infty is not a closed subscheme: any closed subscheme is contained in the pre-image of a closed subscheme from A^n for some n. So I withdraw my example.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.